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Difference between revisions of "Mock AIME 2 2006-2007 Problems/Problem 3"
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== Problem ==
 
== Problem ==
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Let <math>\displaystyle S</math> be the sum of all [[positive integer]]s <math>\displaystyle n</math> such that <math>\displaystyle n^2+12n-2007</math> is a [[perfect square]]. Find the [[remainder]] when <math>\displaystyle S</math> is divided by <math>\displaystyle 1000.</math>
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Let <math>S</math> be the sum of all [[positive integer]]s <math>n</math> such that <math>n^2+12n-2007</math> is a [[perfect square]]. Find the [[remainder]] when <math>S</math> is divided by <math>1000.</math>
 
==Solution==
 
==Solution==
  
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*[[Mock AIME 2 2006-2007/Problem 2 | Previous Problem]]
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*[[Mock AIME 2 2006-2007 Problems/Problem 2 | Previous Problem]]
  
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*[[Mock AIME 2 2006-2007/Problem 4 | Next Problem]]
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*[[Mock AIME 2 2006-2007 Problems/Problem 4 | Next Problem]]
  
 
*[[Mock AIME 2 2006-2007]]
 
*[[Mock AIME 2 2006-2007]]
  
 
[[Category:Intermediate Number Theory Problems]]
 
[[Category:Intermediate Number Theory Problems]]

Revision as of 14:35, 3 April 2012

Problem

Let $S$ be the sum of all positive integers $n$ such that $n^2+12n-2007$ is a perfect square. Find the remainder when $S$ is divided by $1000.$

Solution

If $n^2 + 12n - 2007 = m^2$, we can complete the square on the left-hand side to get $n^2 + 12n + 36 = m^2 + 2043$ so $(n+6)^2 = m^2 + 2043$. Subtracting $m^2$ and factoring the left-hand side, we get $(n + m + 6)(n - m + 6) = 2043$. $2043 = 3^2 \cdot 227$, which can be split into two factors in 3 ways, $2043 \cdot 1 = 3 \cdot 681 = 227 \cdot 9$. This gives us three pairs of equations to solve for $n$:

$n + m + 6 = 2043$ and $n - m + 6 = 1$ give $2n + 12 = 2044$ and $n = 1016$.

$n + m + 6 = 681$ and $n - m + 6 = 3$ give $2n + 12 = 684$ and $n = 336$.

$n + m + 6 = 227$ and $n - m + 6 = 9$ give $2n + 12 = 236$ and $n = 112$.

Finally, $1016 + 336 + 112 = 1464$, so the answer is $464$.

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